A number is
said to be Armstrong if the sum of
the cubes of its digits is equal to the original number.
Example : N = 153 , 13
+ 53 + 33 = 153
// 407 is an Armstrong no.
import
java.util.*;
class Armstrong
{
public static void main(String args[])
{
int
s = 0 , rem ;
Scanner sk = new Scanner(System.in) ;
System.out.print("Enter ur number
: ") ;
int n = sk.nextInt() ;
int temp = n ;
while(n!=0)
{
rem = n % 10 ;
s = s + rem * rem * rem;
n = n / 10 ;
}
if(temp
== s)
{
System.out.println(" The given
number is Armstrong ");
}
else
{
System.out.println(" The given
number is Not Armstrong ");
}
}
}
An Armstrong Number, the sum of power of individual digits is equal to
number itself.
For example 0, 1, 153, 370, 371 and 407 etc
are the Armstrong numbers. That is :
153=1*1*1 + 5*5*5 + 3*3*3
or
1634 = 1*1*1*1 + 6*6*6*6 +
3*3*3*3 + 4*4*4*4
WAP to prove that the sum of
cubes of the digits is equal to the given number.
class Armstrong
{
public static void main(String args[])
{
int num = 1634 ;
int t1 = num , len=0 ;
while(t1 != 0)
{
len = len + 1 ;
t1 = t1 / 10 ;
}
int t2=num , arm=0 ;
while(t2 != 0)
{
int mult = 1 ;
int rem = t2 % 10 ;
for(int i=1 ; i<=len ; i++)
{
mult = mult * rem ;
}
arm = arm + mult ;
t2 = t2/10 ;
}
if(arm==num)
{
System.out.println(" The given
number is Armstrong ");
}
else
{
System.out.println(" The given
number is NOT an Armstrong ");
}
}
}
Wow sir
ReplyDeleteSir I want to know about armstrong
ReplyDelete